calloc() syntax declaration with pointers

Question

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main()
{
    int *ptr,a[6]={1,2,3,4,5,6};
    int i;    
    ptr=(int*)calloc(a[6]*sizeof(int),2);
    for(i=0;i<7;i++)
    {
        printf("\n %d",*ptr+a[i]);
    }
    free(ptr);
    getch();
}

the program works perfectly!! but my questions

Does the declaration of a[6] inside calloc() correct. since the syntax for calloc is ptr_var=(cast_type *)calloc(no_of_blocks,size_of_each_block);.what does the calloc() and a[6] do now.
What does *ptr takes value because ptr doesnt get assigned any address.
In printf does *ptr needs there.then what does it do.

This is pretty meaningless; there is no `a[6]`, and you're supply the size as part of the wrong argument to `calloc`. — Oliver Charlesworth, Oct 01 '14 at 19:38
[Don't cast the result of malloc (and friends)](http://stackoverflow.com/q/605845) — Deduplicator, Oct 01 '14 at 19:39
"the program works perfectly" - are you *sure* about that? Your code invokes *undefined behavior* in the first `calloc` invoke by dereferencing an array one-element-past its last addressable element. Your program is ill-formed. If it "works", it isn't perfectly, its by sheer luck (all of it bad). — WhozCraig, Oct 01 '14 at 19:41
ive tried without a cast.it works on turbo c. the link ishttp://www.vdesignourweb.com/c/c_dynamic_memory.html — t72belt90, Oct 01 '14 at 19:45

chux - Reinstate Monica · Answer 1 · 2014-10-01T19:52:08.637

ptr=(int*)calloc(a[6]*sizeof(int),2); is amiss.
The (int*) cast is not needed.
The a[6] is the value of a[6] which is outside a valid index range of 0 ...5.
Instead the number of elements and the size of size of *ptr are needed.
calloc() will zero-fill the memory assigned to ptr.
```
size_t n = TBD;  // Somehow determine the desired number of `int`s for `ptr`.
ptr = calloc(n, sizeof *ptr);
```
ptr does get a value. To see its value
```
printf("%p\n", (void*) ptr);
```

To print the values of the elements of ptr:

size_t n =  TBD;  // n is the number of elements.
for (size_t i=0; i<n; n++) 
  printf("%d\n", ptr[i]);

a does not need memory allocation as that was done in its declaration. To see its values

    for (size_t i=0; i<sizeof a/sizeof a[0]; i++) 
      printf("%d\n", a[i]);

Haris · Accepted Answer · 2014-10-01T20:09:49.203

3

1.does the declaration of a[6] inside calloc() correct. since the syntax for calloc is

No, its not correct. The array a has 6 elements with indices 0,1,2,3,4,5. but no 6.

2.what does *ptr takes value because ptr doesnt get assigned any address.

ptr does get an address. You are assigning address to ptr when you make the call to calloc() and store the returning address to it. As *ptr is a pointer to an int, it will take integer as a value.

3.in printf does *ptr needs there.then what does it do.

Since you have not stored any value in that allocated memory that ptr points to, it will be all 0's

edited Oct 01 '14 at 20:09

answered Oct 01 '14 at 19:42

Haris

12,120
6
43
70

@haris@chuxok now a[6] should not present inside calloc(). right! – t72belt90 Oct 01 '14 at 20:11
according to your code `a[0] = 1, a[1] = 2, a[2] = 3, a[3] = 4, a[4] = 5, a[5] = 6` now you choose what should be inside calloc() – Haris Oct 01 '14 at 20:12
no my doubt is whether a[6] will impact calloc() or not withot it. – t72belt90 Oct 01 '14 at 20:24
a[6] must present or not – t72belt90 Oct 01 '14 at 20:25
@hariswe will keep it as a[5] itself.again my question is array present inside calloc() is right or not – t72belt90 Oct 01 '14 at 20:38

calloc() syntax declaration with pointers

2 Answers2